The number of values of x in `[0, 4 pi]` satisfying the inequation `|sqrt(3)"cos" x - "sin"x|ge2`, is

Options

A. 0
B. 2
C. 4
D. 8

Explanation

To solve the inequality \( |\sqrt{3} \cos x - \sin x| \geq 2 \) for \( x \) in the interval \( [0, 4\pi] \), we can follow these steps:

### Step 1: Rewrite the Inequality
We start with the inequality:
\[ |\sqrt{3} \cos x - \sin x| \geq 2 \]

### Step 2: Break Down the Absolute Value
This absolute value inequality can be split into two cases:
1. \( \sqrt{3} \cos x - \sin x \geq 2 \)
2. \( \sqrt{3} \cos x - \sin x \leq -2 \)

### Step 3: Solve the First Case
For the first case:
\[ \sqrt{3} \cos x - \sin x \geq 2 \]
Rearranging gives:
\[ \sqrt{3} \cos x \geq \sin x + 2 \]

### Step 4: Solve the Second Case
For the second case:
\[ \sqrt{3} \cos x - \sin x \leq -2 \]
Rearranging gives:
\[ \sqrt{3} \cos x \leq \sin x - 2 \]

### Step 5: Divide by 2
To simplify the expressions, we can divide the entire inequality by 2:
1. For the first case:
\[ \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x \geq 1 \]
2. For the second case:
\[ \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x \leq -1 \]

### Step 6: Use Trigonometric Identities
We can express \( \frac{\sqrt{3}}{2} \) and \( \frac{1}{2} \) in terms of cosine and sine of \( \frac{\pi}{6} \):
\[ \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x = \cos\left(\frac{\pi}{6}\right) \cos x - \sin\left(\frac{\pi}{6}\right) \sin x \]
This can be rewritten using the cosine of a sum:
\[ \cos\left(x + \frac{\pi}{6}\right) \]

### Step 7: Rewrite the Inequalities
Now we have:
1. \( \cos\left(x + \frac{\pi}{6}\right) \geq 1 \)
2. \( \cos\left(x + \frac{\pi}{6}\right) \leq -1 \)

### Step 8: Analyze the Cosine Function
The cosine function equals 1 at:
\[ x + \frac{\pi}{6} = 2n\pi \quad (n \in \mathbb{Z}) \]
Thus:
\[ x = 2n\pi - \frac{\pi}{6} \]

The cosine function equals -1 at:
\[ x + \frac{\pi}{6} = (2n + 1)\pi \quad (n \in \mathbb{Z}) \]
Thus:
\[ x = (2n + 1)\pi - \frac{\pi}{6} \]

### Step 9: Find Values in the Interval
Now we need to find values of \( x \) in the interval \( [0, 4\pi] \).

1. For \( x = 2n\pi - \frac{\pi}{6} \):
- \( n = 0: x = -\frac{\pi}{6} \) (not valid)
- \( n = 1: x = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6} \) (valid)
- \( n = 2: x = 4\pi - \frac{\pi}{6} = \frac{24\pi}{6} - \frac{\pi}{6} = \frac{23\pi}{6} \) (valid)

2. For \( x = (2n + 1)\pi - \frac{\pi}{6} \):
- \( n = 0: x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \) (valid)
- \( n = 1: x = 3\pi - \frac{\pi}{6} = \frac{18\pi}{6} - \frac{\pi}{6} = \frac{17\pi}{6} \) (valid)
- \( n = 2: x = 5\pi - \frac{\pi}{6} = \frac{30\pi}{6} - \frac{\pi}{6} = \frac{29\pi}{6} \) (not valid)

### Step 10: Count the Valid Solutions
The valid solutions found are:
1. \( \frac{5\pi}{6} \)
2. \( \frac{11\pi}{6} \)
3. \( \frac{17\pi}{6} \)
4. \( \frac{23\pi}{6} \)

Thus, the total number of solutions is **4**.